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Suppose $h$ is bounded by one in absolute value on a punctured neighbourhood of zero. Suppose also that there are positive sequences $p_n,q_n\to 0$ with $h(p_n)=1$ and $h(q_n)=0$. Extend $F_a(x)=|x{|}^{a}h(x)$ by $F_a(0)=0$. ThenWhen the derivative exists, it equals zero. For continuity, the bound $|F_a(x)|\le |x{|}^a$ gives the epsilon-delta definition of continuity with $\delta ={\epsilon }^{1/a}$ when $a>0$. If $a\le 0$, the values $F_a(p_n)=p_n^a$ remain at least one for sufficiently large $n$, contradicting continuity at zero.

For differentiability, the difference quotient is bounded in absolute value by $|x{|}^{a-1}$, so it tends to zero when $a>1$. Conversely its values along $q_n$ are zero, forcing any possible derivative to be zero. Its values along $p_n$ equal $p_n^{a-1}$ and fail to tend to zero when $a\le 1$.

For example, $h(x)=\sin (\pi \sin (1/x))$ has the required positive sequences $p_n=1/(2\pi n+\pi /6)$ and $q_n=1/(2\pi n)$. This separates the suppression needed for continuity from the stronger suppression needed for differentiability.

A continuous function sends sufficiently nearby function inputs to nearby function outputs. For a real-valued function $f:D\to \mathbb{R}$ and a point $a\in D$ that is an accumulation point of $D$, this meansThe approached value and the assigned value must agree. Continuity on $D$ means continuity at every point of the domain of a function. At an endpoint of an interval, only approaches inside that domain of a function count. The epsilon-delta definition of continuity below also applies at an isolated point, where continuity holds automatically.^[ref]

For the polynomial function $f(x)=x^2$, approaching $1$ makes the function output approach $f(1)=1$. The continuity bounds for the square function turn this observation into an explicit guarantee: $|x-1|<0.2$ implies $|x^2-1|<0.6$. This is useful whenever a small error in an function input must produce a controlled error in an function output. The required input tolerance can depend on the point and on the chosen output tolerance.

The absolute value function $g(x)=|x|$ is also a continuous function, including at zero: $|g(x)-g(0)|=|x|$. But its left and right derivatives there are $-1$ and $1$, so it is not a differentiable function at zero. Continuity controls values without requiring a tangent line. Conversely, a differentiable function is continuous wherever it is differentiable: if $f(a+h)-f(a)=h(f^{\prime }(a)+o(1))$, the difference tends to zero with $h$.

Several useful consequences reuse this one local idea. Addition, multiplication and function composition preserve continuity; a quotient of functions is continuous where its denominator is nonzero. On an interval, the intermediate value theorem turns continuity into a guarantee that no intermediate function output is skipped. On a compact set, the extreme value theorem guarantees attained maximum and minimum values. Those extra hypotheses matter: continuity alone on an unbounded domain of a function does not provide a global bound or an attained global maximum.

A rational function such as $x\mapsto 1/x$ is continuous on its domain of a function $\mathbb{R}\setminus \{0\}$. Zero is excluded from that domain of a function; the missing point prevents a continuous extension there, but it does not make the existing map discontinuous at one of its actual inputs. Always state the domain of a function before judging continuity.

For $f:D\to \mathbb{R}$ and $a\in D$, continuity at $a$ is equivalent to the following condition: every sequence $(x_n)$ in $D$ with $x_n\to a$ satisfies $f(x_n)\to f(a)$. This is useful when a numerical method or a proof approaches a point through successively refined function inputs: applying a continuous function preserves the intended limit.

For the forward direction, choose $\delta$ from the epsilon-delta definition of continuity. A convergent sequence eventually lies in that input window, so its function outputs eventually lie in the requested output band. Conversely, if continuity fails, there is some ${\epsilon }_0>0$ for which every input window contains an allowed $x$ with $|f(x)-f(a)|\ge {\epsilon }_0$. Choosing such an $x_n$ within $1/n$ of $a$ gives $x_n\to a$ while $f(x_n)$ does not tend to $f(a)$. This converse relies on the countable shrinking windows available in metric spaces; a general topological space needs a different formulation.

A function is discontinuous at a point in its domain of a function when it fails the epsilon-delta definition of continuity there. Two basic failures are an incorrect assigned value and unequal one-sided limits. They demand different repairs.

For a removable discontinuity, considerThe limit at zero is $0$, but $r(0)=1$. Replacing that one value by $0$ gives the continuous square function. The sequence $x_n=1/n$ makes the mismatch visible: $r(x_n)=1/n^2\to 0\ne r(0)$. A removable discontinuity can also describe an expression with a missing value, in which case the question is whether a continuous extension to the missing input exists.

For a jump discontinuity, take $j(x)=0$ for $x<0$ and $j(x)=1$ for $x\ge 0$. The left and right one-sided limits at zero are $0$ and $1$. No assignment to $j(0)$ can equal both, so changing only the point value cannot make the function continuous there. Contrast both examples with the corner of $|x|$: that function is continuous at zero even though its derivative is absent.

Continuity. The function is continuous at $x_0$ when ${\lim }_{x\to x_0}f(x)=f(x_0)$. Equivalently, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ within the domain of a function implies $|f(x)-f(x_0)|<\epsilon$; this is the epsilon-delta definition of continuity.

Differentiability. At an interior point $x_0$, the difference quotient must have a finite two-sided limit of a function:This defines differentiability and the derivative. At an endpoint one may instead state a one-sided definition relative to the interval. Differentiability implies continuity, because the increment is $h$ times a quotient with finite limit, but continuity alone does not guarantee differentiability.

For statement (1), continuity at the interior point meansExplicitly, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$ for points in the domain. This is the epsilon-delta definition of continuity.

For statement (2), differentiability means the two-sided difference quotient has a finite limit:The same finite value must be obtained from positive and negative $h$. Differentiability implies continuity, whereas the converse need not hold.

The bound $|\cos (1/x)|\le 1$ gives $|f(x)-f(0)|\le |x{|}^p$ when $x\ne 0$. If $p>0$, given any $\epsilon >0$, choose $\delta ={\epsilon }^{1/p}$. Then $0<|x|<\delta$ impliesthe inequality also holds at $x=0$. This proves continuity using the epsilon-delta definition of continuity.

For necessity, take the positive sequence $x_n=1/(2\pi n)\to 0$. Then $\cos (1/x_n)=1$, so $f(x_n)=(2\pi n{)}^{-p}$. For $p=0$ this equals $1$; for $p<0$ it tends to $+\infty$. In neither case does it tend to $f(0)=0$, so continuity fails. ConsequentlyThis is an instance of the general continuity and differentiability thresholds for a damped oscillation.

The epsilon-delta definition of continuity says that $f$ is continuous at $a$ ifHere $a$ belongs to the domain of $f$, and the implication is required for every $x$ in that domain. The number $\delta$ may depend on $\epsilon$ and $a$, but must work for all such $x$ simultaneously.

The function $f$ is continuous at $x_0$ ifThe choice of $\delta$ may depend on $\epsilon$ and $x_0$, but must work for every $x$ in that neighbourhood. This is the epsilon-delta definition of continuity.

For every $\epsilon >0$ there must be a $\delta >0$ such thatThe choice of $\delta$ may depend on $x_0$ and $\epsilon$. This is the epsilon-delta definition of continuity.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.